To show that $| \sin x | \leq | x |$ for all $x \in \mathbb{R}$ using the Mean Value Theorem (MVT), we'll proceed as follows:

1. Understanding the MVT:

The Mean Value Theorem (MVT) states that if a function $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$, then there exists at least one point $c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In our case, we will apply this to the function $f(x) = \sin x$ on the interval $[0, x]$ for $x \neq 0$.

2. Applying the MVT:

Let's consider the function $f(x) = \sin x$. We know that $\sin x$ is continuous and differentiable on $\mathbb{R}$, so we can apply the MVT to $f(x)$ on the interval $[0, x]$ for $x \neq 0$.

By the MVT, there exists some $c \in (0, x)$ such that:

$\frac{\sin x - \sin 0}{x - 0} = \cos(c)$

Since $\sin 0 = 0$, this simplifies to:

$\frac{\sin x}{x} = \cos(c)$

Thus, we have:

$\sin x = x \cos(c)$

3. Bounding $|\sin x|$:

From the equation $\sin x = x \cos(c)$, we take the absolute value of both sides:

$|\sin x| = |x \cos(c)|$

Since $|\cos(c)| \leq 1$ for all $c$, it follows that:

$|\sin x| \leq |x|$

4. Conclusion:

We have shown that $|\sin x| \leq |x|$ for all $x \neq 0$ using the MVT. For $x = 0$, $\sin 0 = 0$, so the inequality holds trivially.

Thus, the inequality $|\sin x| \leq |x|$ holds for all $x \in \mathbb{R}$.

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Would you like to explore further details or need more clarifications? Here's a set of related questions:

1. How can we extend this result to other trigonometric functions using the MVT?
2. What is the connection between the MVT and the Taylor series of $\sin x$?
3. How does this result compare with the approximation $\sin x \approx x$ for small $x$?
4. Can the MVT be applied to prove other inequalities like $\cos x \leq 1$?
5. What are the assumptions required for the MVT to be applicable?

Tip: The MVT provides a powerful tool for proving inequalities in calculus, especially when working with differentiable functions like trigonometric functions.